Max Throw

Stage: 5 Short Challenge Level:

Suppose that the particle is projected from a height $H$ above the ground at speed $V$ at an angle $\alpha$ to the $x$-axis, with $x$ measuring the horizontal distance travelled and $y$ measuring the vertical distance travelled.

The particle intersects the $x$-axis when the $y$-coordinate is zero. This is when
$$
t = \frac{V\sin(\alpha) \pm \sqrt{V^2\sin^2(\alpha) +2gH}}{g}\;.
$$

The particular case $H=0$
The square root simplifies giving the point of intersection as
$$
(x, y) = \left(\frac{2V^2}{g}\cos(\alpha)\sin(\alpha), 0\right) = \left(\frac{2V^2}{g}\sin(2\alpha), 0\right)\,,
$$
where the second equality makes use of a trig identity. The $x$-value $V\sin(2\alpha)$ is maximised when $2\alpha=90^\circ$. Thus the optimal angle of projection is $45^\circ$, for any initial velocity.

This looks complicated to differentiate so I tried a numerical solution using a spreadsheet.This produced an optimal angle of $40.4^\circ$ (3sf).

It seems clear that this angle will be dependent on the initial speed, but to check I calculate the optimum angle numerically for a large initial speed of $100\mathrm{ms}^{-1}$. This produces an optimium of $44.9^\circ$ (3sf) which is close to $45^\circ$, as we might intuitively expect.

Extension:

You might like to investigate this further. Here are some starting points.

Note that the expression for the derivative is:
$$
\frac{g}{V^2}\frac{dx}{d\alpha} = \cos(2\alpha)-\sin(\alpha)\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}+\cos^2\alpha\sin\alpha \frac{1}{\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}}\;.
$$
For an optimum we can set the left hand side to zero. This gives us
$$
\sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right) + \sin(\alpha)\cos(2\alpha)\left(1 +\frac{2gH}{V^2\sin^2(\alpha)}\right)^{\frac{1}{2}}=0\;.
$$

If $X(\alpha) \equiv \frac{2gH}{V^2\sin^2(\alpha)}$ is small then we can expand to give
$$
\sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right)+ \sin(\alpha)\cos(2\alpha)\left(1+\frac{1}{2}\frac{2gH}{V^2\sin^2(\alpha)}+\mathcal{O}(X^2)\right)=0\;.
$$
In principle that can now be turned into a polynomial expression in $\cos(\alpha)\ldots$

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